Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes.

Since the circle touches both the axes, the centre of the circle must lie on the line x+y=0 in the second quadrant, which is equidistant from both the co-ordinate axes. Let the centre be (-a,a) [x is negative and y is positive because the centre lies in the second quadrant] The radius of the circle will be the distance of the centre from either of the axes =a The equation of the family of circles will then be (x+a)²+(y-a)²=a²(i) Now, differentiate with respect to x and let dy/dx=y'
2(x+a)+2(y-a)y'=0
⇒x+yy'+a(1-y')=0
⇒a=(x+yy')/(y'-1)Substitute this value of a in (i)
(x+(x+yy')/(y'-1))²+(y-(x+yy')/(y'-1))²=((x+yy')/(y'-1))²
⇒(xy'-x+x+yy')²+(y(y'-1)-x-yy')²=(x+yy')²
⇒(xy'+yy')²+(y-x)²=(x+yy')²
⇒x²y'²+y²y'²+2xyy'²+x²+y²-2xy=(x+yy')²
⇒x²y'²+y²y'²+2xyy'²+x²+y²-2xy=x²+y²y'²+2xyy'
⇒y'²(x²+2xy)-2xyy'-y²+2xy=0The above is the differential equation for the given question.